# An object is thrown vertically from a height of 5 "m" at 3 "m/s". How long will it take for the object to hit the ground?

Mar 4, 2016

$0.7 \quad \text{s}$

#### Explanation:

"Thrown vertically" could mean either upwards or downwards. I am assuming it to be downwards.

Step 1: Understand the problem.

An object is under free fall. It is accelerating at $g = 9.81 \quad {\text{m/s}}^{2}$ downwards. It is given an initial velocity of $3 \quad \text{m/s}$ downwards.

Step 2: Identify the relevant equations.

To find displacement, $s$, under constant acceleration, $a$, and initial velocity, ${v}_{0}$, after time, $t$, the following equation is applicable:

$s = {v}_{0} t + \frac{1}{2} a {t}^{2}$

To get the time as a function of the distance traveled, make $t$ as the subject of formula. Begin by writing the quadratic equation in standard form.

${t}^{2} + \frac{2 {v}_{0}}{a} t - \frac{2 s}{a} = 0$

Subsequently, complete the square.

${\left(t + \frac{{v}_{0}}{a}\right)}^{2} - \frac{2 s}{a} = {\left(\frac{{v}_{0}}{a}\right)}^{2}$

${\left(t + \frac{{v}_{0}}{a}\right)}^{2} = \frac{{v}_{0}^{2}}{{a}^{2}} + \frac{2 s}{a}$

$= \frac{{v}_{0}^{2} + 2 a s}{{a}^{2}}$

Reject the negative square root and take the positive one. This is because we only restrict ourselves to $t > 0$.

$t + \frac{{v}_{0}}{a} = \frac{\sqrt{{v}_{0}^{2} + 2 a s}}{a}$

$t = \frac{\sqrt{{v}_{0}^{2} + 2 a s} - {v}_{0}}{a}$

Step 3: Set up the coordinate system.

I let the downwards direction be positive. The ground level is arbitrarily set to be the origin (i.e. $s = 0$). This is up to your preferences.

Step 4: Identiy the parameters.

In this question,

• $a = g = 9.81 \quad {\text{m/s}}^{2}$

• ${v}_{0} = 3 \quad \text{m/s}$

• $s = 5 \quad \text{m}$

Step 5: Plug into the equation.

$t = \frac{\sqrt{{v}_{0}^{2} + 2 a s} - {v}_{0}}{a}$

= frac{sqrt{(3 quad "m/s")^2 + 2(9.81 quad "m/s"^2)(5 quad "m")} - (3 quad "m/s")}{9.81 quad "m/s"^2}

$= 0.749 \quad \text{s}$

Now earlier, I assume that the object has an initial velocity of $3 \quad \text{m/s}$ downwards. The formula also works if the initial velocity is $3 \quad \text{m/s}$ upwards. In that case, everything remains the same, except

• ${v}_{0} = - 3 \quad \text{m/s}$

Plugging in the values again

$t = \frac{\sqrt{{v}_{0}^{2} + 2 a s} - {v}_{0}}{a}$

= frac{sqrt{(-3 quad "m/s")^2 + 2(9.81 quad "m/s"^2)(5 quad "m")} - (-3 quad "m/s")}{9.81 quad "m/s"^2}

$= 1.361 \quad \text{s}$